Problem: How many unique ways are there to arrange the letters in the word PRIOR?
Solution: Let's try building the arrangements (or permutations) letter by letter. The word is $5$ letters long: _ _ _ _ _ Now, for the first blank, we have $5$ choices of letters to put in. After we put in the first letter, let's say it's I, we have $4$ blanks left. I _ _ _ _ For the second blank, we only have $4$ choices of letters left to put in. So far, there were $5 \cdot 4$ unique choices we could have made. We can continue in this fashion to put in a third letter, then a fourth, and so on. At each step, we have one fewer choice to make, until we get to the last letter, and there's only one we can put in. Using this method, the total number of arrangements is $5\cdot4\cdot3\cdot2\cdot1 = 120.$ Another way of writing this is $5!$, or $5$ factorial, but this isn't quite the right answer. Using the above method, we assumed that all the letters were unique. But they're not! There are $2$ Rs, so we're counting every permutation multiple times. So every time we have these $2$ permutations: IRORP IRORP We actually should have only one permutation: IRORP Notice that we've overcounted our arrangements by $2!$. This is not a coincidence! This is exactly the number of ways to permute $2$ objects, which we were doing with the non-unique Rs. To address this overcounting, we need to divide the number of arrangements we counted before by $2!$. When we divide the number of permutations we got by the number of times we're overcounting each permutation, we get $ \dfrac{5!}{2!} = \dfrac{120}{2} = 60$